# Why $z \in \overline{A}$? [closed]

In the Picture blew: The paper can be downloaded here. Why $z \in \overline{A}$? Thanks.

A point $x$ of a space $X$ is called $G_\omega$-separated from a subset $Y$ of $X$ if there is a closed $G_\omega$ -set $P$ in $X$ such that $x \in P$ and the sets $Y$ and $P$ are disjoint.

## closed as off topic by Ramiro de la Vega, Lee Mosher, Ryan Budney, Daniel Moskovich, George LowtherFeb 26 '13 at 12:14

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

• Downvoted, although I haven't voted to close (yet). – Yemon Choi Feb 25 '13 at 16:28
• Given the existing votes to close, I suggest that discussion (about closing or keeping open this question) takes place at tea.mathoverflow.net/discussion/1539/… – Yemon Choi Feb 25 '13 at 16:40
• @Yemon Choi: You are not very welcome. It seems that you only can do this, what other could you do? – Paul Feb 26 '13 at 3:45
• @Luke: perhaps you would get a better reception if you gave some explanation of why you have posted so many questions in general topology, often in complicated areas but with the questions usually seeming to have quick answers – Yemon Choi Feb 26 '13 at 6:56
• Speaking for myself: I'm afraid I don't react well to someone who uses to different names on different sites, and in neither case is upfront about their motivation for questions or previous experience – Yemon Choi Feb 26 '13 at 6:58

Since $Z$ is regular, for any open neighborhood $U$ of $z$ there is an open $V$ such that $z\in V$ and $\overline{V}\subseteq U$ (apply regularity to $z$ and the complement of $U$). Do this countably many times and take intersection, and you'll find that for any open neighborhood $U$ of $z$ there is a closed $G_\omega$-set $P$ such that $z\in P\subseteq U$.
If follows easily that if $z$ is not $G_\omega$-separated from $A$ (that is, if (b) fails), then $z$ must be in the closure of $A$.
• @Todd Eisworth: How could you see that $\cap V_i$ is closed as a closed $G_\omega$-set $P$ of $z$? – Paul Feb 26 '13 at 3:50
• That is, you make sure $V_{i+1}\subseteq\overline{V}_{i+1}\subseteq V_i$, and then $\cap V_i = \cap \overline{V}_i$ is closed and a $G_\omega$ set. – Todd Eisworth Feb 26 '13 at 4:01